Question

At 27°C, 100 mL of 0.5 M HCl is mixed with 100 mL of 0.4 M NaOH solution. To this resultant solution, 800 mL of distilled water is added. What is the pH of the final solution? 

• A. \( 12.0 \)
• B. \( 1.3 \)
• C. \( 1.3 \)
• D. \( 1.0 \)

Hint:

When a strong acid (HCl) and a strong base (NaOH) are mixed, find the limiting reagent and determine the excess moles remaining. Then, use the total volume to calculate the final concentration and determine the pH.

Answer 1

The Correct Option is B

Step 1: Calculate the Moles of HCl and NaOH 
\[ \text{Moles of HCl} = M \times V = (0.5 \text{ M}) \times (0.1 \text{ L}) = 0.05 \text{ moles} \] \[ \text{Moles of NaOH} = M \times V = (0.4 \text{ M}) \times (0.1 \text{ L}) = 0.04 \text{ moles} \]

 Step 2: Determine the Excess Acid or Base 
Since HCl and NaOH react in a 1:1 ratio: \[ \text{Remaining HCl} = 0.05 - 0.04 = 0.01 \text{ moles} \] 

Step 3: Calculate the Final Concentration of \( H^+ \) 
Total volume after mixing: \[ 100 \text{ mL} + 100 \text{ mL} + 800 \text{ mL} = 1000 \text{ mL} = 1 \text{ L} \] \[ [\text{H}^+] = \frac{\text{Remaining moles of } HCl}{\text{Total volume in L}} = \frac{0.01}{1} = 0.01 \text{ M} \]

 Step 4: Calculate the pH 
\[ \text{pH} = -\log [\text{H}^+] \] \[ \text{pH} = -\log (0.01) = 2.0 \] 

Step 5: Verify the Correct Answer 
Thus, the final pH of the solution is \( 2.0 \), which matches Option (2).